Question

it is possible to build a clock in which the tips of the hour hand and the second hand move with the same tangential speed. this is normally never done, however. why?
a. the length of the hour hand would be 720 times smaller than the length of the second hand.
b. the length of the hour hand would be 720 times greater than the length of the second hand.
c. the hour hand and the second hand would have the same length.

Explanation:

Step1: Recall the formula for tangential speed

The formula for tangential speed \(v = r\omega\), where \(r\) is the radius (length of the hand in this case) and \(\omega\) is the angular speed.

Step2: Find the angular speeds of the hour and second hands

The second - hand makes one full revolution (\(2\pi\) radians) in \(T_{s}=60\ s\). So, \(\omega_{s}=\frac{2\pi}{T_{s}}=\frac{2\pi}{60}\ rad/s\).
The hour - hand makes one full revolution (\(2\pi\) radians) in \(T_{h}=12\times3600\ s\) (since there are 12 hours in a 12 - hour cycle and 3600 seconds in an hour). So, \(\omega_{h}=\frac{2\pi}{T_{h}}=\frac{2\pi}{12\times3600}\ rad/s\).

Step3: Set the tangential speeds equal and solve for the ratio of lengths

Since \(v_{h}=v_{s}\), then \(r_{h}\omega_{h}=r_{s}\omega_{s}\).
Substitute \(\omega_{h}\) and \(\omega_{s}\):

$$
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$$

So, \(r_{h} = 720r_{s}\)

Answer:

B. The length of the hour hand would be 720 times greater than the length of the second hand.